Chapter 1

Angular Momentum in Quantum Mechanics

1.1 Classical vectors vs quantum operators

1.1.1 Relevant definitions and notation In classical physics, the angular momentum L of a particle is given by the cross product of its position vector r and momentum vector p: L = r p. (1.1) ⇥ In quantum physics, we can construct an analogous operator by defining the ”vector” oper- ators rˆ and pˆ as follows: rˆ =ˆx e +ˆx e +ˆx e 1 1 2 2 3 3 (1.2) pˆ =ˆp1e1 +ˆp2e2 +ˆp3e3 where i =1, 2, 3correspondtox, y, z,resp.,andtheei are ortho-normal unit vectors in the 3-dimensional vector space. The ”hat” (ˆ) symbols signify that these are operators, and the bold type signifies that it is a construction of operators with unit vectors, which we will refer to as vector operators. This notation allows us to perform the classical vector operations, such as dot- and cross-products. In particular, by maintaining the classical equation for the angular momentum vector, we are able to write the Lˆi component of the angular momentum operator given in Eq. 1.1 in the following way:

Lˆ =ˆx pˆ xˆ pˆ 1 2 3 3 2 Lˆ =ˆx pˆ xˆ pˆ (1.3) 2 3 1 1 3 Lˆ =ˆx pˆ xˆ pˆ 3 1 2 2 1 and the orbital angular momentum vector operator is Lˆ = Lˆ1e1 + Lˆ2e2 + Lˆ3e3. Eq. 1.3 can be more compactly written as:

Lˆi = "ijkxˆjpˆk (1.4)

5 6 CHAPTER 1. ANGULAR MOMENTUM IN QUANTUM MECHANICS

where "ijk is the three-dimensional Levi-Civita symbol, and we employ the Einstein summa- tion convention, i.e. repeated indices imply a sum over that index. This notation will be used in other parts of these notes.

1.1.2 Quantum vector operations In order to build up a formalism using our quantum vector operators, we need to examine some of their important properties. While the classical position and momentum xi and pi commute, this is not the case in quantum mechanics. The commutation relations between position and momentum operators is given by:

[ˆxi, xˆj]=0, [ˆpi, pˆj]=0, [ˆxi, pˆj]=i~ij, (1.5) where ij is the Kronecker delta symbol. It should be noted that the ei unit vectors commute with all operators. To see how the vector operators behave under dot and cross products, we can apply them in the normal way: uˆ vˆ =ˆu vˆ i i (1.6) (uˆ vˆ· ) = " uˆ vˆ ⇥ i ijk j k Unlike with classical vectors, these are operators, and their order can, in general, not be changed. For the case of the dot-product, switching the order of the vector operators results in: uˆ vˆ =[ˆu , vˆ ]+ˆv uˆ i i i i (1.7) · = vˆ uˆ +[ˆu , vˆ ] · i i Contrast this with the classical dot product result, where u v = v u.Itisnowclearthat the classical result holds true for commuting operators, but· not in· the general case where the operators may not commute. Note the special case of uˆ2 = uˆ uˆ, which does obey the classical dot product relationship. · Similarly, the cross product can we written as:

(uˆ vˆ) = " ([ˆu , vˆ ]+ˆv uˆ ) ⇥ i ijk j k k j = "ikjvˆkuˆj + "ijk [ˆuj, vˆk] (1.8) = (vˆ uˆ) + " [ˆu , vˆ ] ⇥ i ijk j k which, again, contrasts with the classical result u v = v u in the case of non-commuting operators. ⇥ ⇥ If we look at the special cases rˆ rˆ and pˆ pˆ we see that: ⇥ ⇥ 1 (rˆ rˆ) =(pˆ pˆ) = " =0 (1.9) ⇥ i ⇥ i 2 ijk jk where the last equality is obvious due to the contradictory requirements of "ijk and jk to be non-zero. 1.1. CLASSICAL VECTORS VS QUANTUM OPERATORS 7

In the case where uˆ = rˆ and vˆ = pˆ,Eq.1.7andEq.1.8resultin: rˆ pˆ = pˆ rˆ +3i ~ (1.10) rˆ · pˆ = p·ˆ rˆ ⇥ ⇥ The cross product between rˆ and pˆ obeys classical rules, however the dot product does not.

1.1.3 Properties of angular momentum A simple analysis can be done to show:

(uˆ vˆ)† = vˆ† uˆ† · · (1.11) (uˆ vˆ)† = vˆ† uˆ† ⇥ ⇥ This property can be used to show that Lˆ is Hermitian (given that rˆ and pˆ are Hermitian):

Lˆ† =(rˆ pˆ)† = pˆ rˆ = Lˆ. (1.12) ⇥ ⇥ It is simple to determine that Lˆ is orthogonal to both rˆ and pˆ,justasitisclassically:

rˆ Lˆ =ˆxiLˆi, · = " xˆ xˆ pˆ =0, ijk i j k (1.13) pˆ Lˆ =ˆp Lˆ , · i i = "ijkpˆixˆjpˆk =0.

The last step of each equation is true due to the sign change of "ijk under a swap of indices, and the commutation properties of position and momentum operators. To see how angular momentum commutes with position or momentum, we can use Eq. 1.4 to write: [Lˆ , xˆ ]=" xˆ pˆ xˆ " xˆ xˆ pˆ i j ikl k l j ikl j k l = "ikl xˆk (ˆxjpˆl i~jl) xˆjxˆkpˆl { } (1.14) = i~"ikjxˆk + "ikl (ˆxkxˆjpˆl xˆjxˆkpˆl) = i~"ijkxˆk The treatment for momentum is very similar, and in summary we obtain the equations: ˆ [Li, xˆj]=i~"ijkxˆk ˆ (1.15) [Li, pˆj]=i~"ijkpˆk These equations are said to mean that rˆ and pˆ are vectors under rotations. Using Eq. 1.15 it is possible to derive the commutation relations between di↵erent compo- nents of Lˆ:

[Lˆi, Lˆj]=Lˆi"jklxˆkpˆl "jklxˆkpˆlLˆi ˆ ˆ = "jkl(ˆxkLi + i~"ikuxˆu)ˆpl "iklxˆkpˆlLi ˆ ˆ = i~"jkl"ikuxˆupˆl + "jklxˆk(ˆplLi + i~"ilvpˆv) "iklxˆkpˆlLi = i~("kjl"kiuxˆupˆl + "lkj"livxˆkpˆv) (1.16) = i~ [(ijlu ilju)ˆxupˆl +(ikjv ijkv)ˆxkpˆv] = i~(ijxˆlpˆl xˆjpˆi +ˆxipˆj ijxˆkpˆk) = i~(ˆxipˆj xˆjpˆi) 8 CHAPTER 1. ANGULAR MOMENTUM IN QUANTUM MECHANICS where in the fourth step we made use of the identity " " = . ijk ilm jl km jm kl We can write the result in Eq. 1.16 explicitly as:

ˆ ˆ ˆ [Lx, Ly]=i~Lz, ˆ ˆ ˆ [Ly, Lz]=i~Lx, (1.17) ˆ ˆ ˆ [Lz, Lx]=i~Ly, and which can be compactly written as:

ˆ ˆ ˆ [Li, Lj]=i~"ijkLk. (1.18)

It is now clear that di↵erent components of the angular momentum do not commute. We can additionally use Eq. 1.18 to straightforwardly show that:

[Lˆ , rˆ2]=0, [Lˆ , pˆ2]=0,,[ˆp , rˆ pˆ]=0,,[Lˆ , Lˆ2]=0, (1.19) i i i · i 2 where in particular the last equation (for Lˆ ) is important. Even though Lˆi does not commute with any other component of Lˆ,itdoescommutewithLˆ2. There is a more elegant way to express the algebra of angular momentum. Using Eq. 1.8 and Eq. 1.18 we can write

(Lˆ Lˆ)i = (Lˆ Lˆ)i + "ijk[Lˆj, Lˆk] ⇥ i~ ⇥ ˆ = 2 "jki"jkmLm, i~ ˆ = 2 (kkim kmik)Lm, (1.20) i~ ˆ ˆ = 2 (Li ikLk), ˆ = i~Li, or, in short Lˆ Lˆ = i~Lˆ. (1.21) ⇥ This result gives an intuitive feel for why angular momentum behaves in an inherently quantum way. A non-zero cross product of a vector with itself is contrary to any classical phenomenom. It is also important to note that, since we only used the general mathematical properties of the cross product of quantum vectors, and the commutation relations of angular momentum operators, Eq. 1.18 and Eq. 1.21 are completely equivalent to each other.

1.2 Angular momentum and central potentials

We now consider the orbital angular momentum in the context of central potentials (poten- tials that depend only on r = prˆ2), such as the Coulomb potential of an electron in the electric field of an atomic nucleus. The Hamiltonian in this case is given by

pˆ2 Hˆ = + V (r),r= prˆ2. (1.22) 2m 1.3. ALGEBRA OF ANGULAR MOMENTUM 9

We can write this Hamiltonian in position space using pˆ = i~ ,orpˆ2 = ~2 2,where 2 is the Laplace operator. In spherical coordinates, this is r r r 1 @2 1 1 @ @ 1 @2 pˆ2 = ~2 r + sin ✓ + (1.23) r @r2 r2 sin ✓ @✓ @✓ sin2 ✓ @2  ✓ ◆

We can write Lˆ2 in spherical coordinates as follows:

Lˆ2 = ~2(rˆ ) (rˆ ) ⇥r · ⇥r 2 ˆ 1 @ ˆ @ ˆ 1 @ ˆ @ = ~ ✓ + ✓ + sin ✓ @ @ · sin ✓ @ @ 2 h ⇣✓ˆ ˆ @2⌘ @✓ˆ @i h ˆ⇣@✓ˆ 1⌘ @ i ✓ˆ @ˆ @ ˆ ˆ @2 @ˆ @ = ~ 2 ✓ 2 + + + + 2 sin ✓ @ @ @ @✓ sin ✓ @ sin ✓ @ @✓ @✓ @✓ @✓ (1.24) 2 h 1 h@2 cos ✓ @ i @2 ⇣ ⌘ ⇣ ⌘i = ~ 2 2 + + 2 sin ✓ @ sin ✓ @✓ @✓ 2 ⇣ 1 @ @ 1 @2⌘ = ~ sin ✓ + 2 2 sin ✓ @✓ @✓ sin ✓ @ ⇣ ⌘ Note that the angular momentum has no r-dependence. Using Eq. 1.24 and Eq. 1.23, we can rewrite the Hamiltonian as:

~2 @2 Lˆ2 Hˆ = r + + V (r)(1.25) 2mr @r2 2mr2

We can now assemble a list of Hermitian operators (observables) we have discussed:

2 2 2 H,ˆ xˆ1, xˆ2, xˆ3, pˆ1, pˆ2, pˆ3, rˆ , pˆ , Lˆ1, Lˆ2, Lˆ3, Lˆ (1.26) where we include all unique Hermitian operators that include up to squared terms of position or momentum. Typically in atomic physics, the spectrum (energy levels) of an atom is of major interest. We therefore want to only consider observables that commute with the Hamiltonian. This 2 excludes thex ˆi operators (don’t commute with pˆ )andˆpi operators (don’t commute with V (r), in general). The same reasoning eliminates rˆ2 and pˆ2. However, from Eq. 1.25 it is 2 clear that the Lˆi operators and Lˆ commute with H.DuetotheLˆi not commuting with each other, we can only use one of them in our set of operators. The convention for this is Lˆz. Our set of commuting Hermitian operators (observables) is then:

2 H,ˆ Lˆz, Lˆ (1.27)

1.3 Algebra of angular momentum

Hermitian operators Jˆx, Jˆy,andJˆz are said to satisfy the algebra of angular momentum if they obey the commutation relations given by Eq. 1.18: ˆ ˆ ˆ [Ji, Jj]=i~"ijkJk. (1.28) 10 CHAPTER 1. ANGULAR MOMENTUM IN QUANTUM MECHANICS

Keep in mind that this could be orbital angular momentum Lˆi,spinangularmomentumSˆi, or any other set of Hermitian operators that obeys the above commutation relation. As we showed in section 1.1.3, from this algebra it also follows that:

2 [Jˆi, Jˆ ]=0. (1.29)

Define then the following operators:

Jˆ = Jˆx iJˆy. (1.30) ± ± ˆ ˆ Note that these operators are each other’s Hermitian conjugate J+† = J . To understand the behavior of these operators and how they act on the angular momentum operators, we can do some simple manipulations:

ˆ ˆ ˆ2 ˆ2 ˆ ˆ ˆ2 ˆ2 ˆ J J = Jx + Jy i[Jx, Jy]=Jx + Jy ~Jz, (1.31) ± ⌥ ⌥ ± and so the commutator is ˆ ˆ ˆ [J+, J ]=2~Jz. (1.32) In addition, we can see the following are all equal:

ˆ2 ˆ2 ˆ2 ˆ2 ˆ ˆ ˆ ˆ ˆ ˆ 1 ˆ ˆ ˆ ˆ J Jz = Jx + Jy = J+J ~Jz = J J+ + ~Jz = (J+J + J J+)(1.33) 2

Since we also know that the individual components commute with Jˆ2 we know that the same must be true for Jˆ : ± [Jˆ , Jˆ2]=0. (1.34) ± Lastly, we calculate the commutation with Jˆz:

[Jˆz, Jˆ ]=[Jˆz, Jˆx] i[Jˆz, Jˆy] ± ˆ ± ˆ = i~Jy i( i~)Jx (1.35) ± = ~Jˆ ± ±

2 Since each Jˆi and Jˆ commute, we can simultaneously diagonalize them. However, the di↵erent Jˆi do not commute with each other, so we have to choose one. Typically, this is Jˆz. The simultaneous eigenstates are denoted by j, m ,andtheyformanorthonormalbasis.So | i far, the only restriction on j and m is j, m R. 2 We label the states in such a way that the following relations apply:

Jˆ2 j, m = ~2j(j +1) j, m ˆ | i | i (1.36) Jz j, m = ~m j, m | i | i The reason why we choose to define the values using j(j +1)(asopposedto j2)isforfuture algebraic convenience, as we will see. 1.3. ALGEBRA OF ANGULAR MOMENTUM 11

Since it is a requirement that j(j +1) 0, it is necessary that either j 1prj 0. However, since the only thing that matters is the eigenvalue, we can just choosej 0. To see how the operators Jˆ act on j, m we will check the eigenvalue of Jˆ2 after acting on ± astate: | i Jˆ2(Jˆ j, m )=Jˆ Jˆ2 j, m = ~2j(j +1)(Jˆ j, m ). (1.37) ±| i ± | i ± | i From this it is clear that Jˆ is still an eigenstate of Jˆ2 while keeping the eigenvalue j ± unchanged. However, it is still unknown if it changes the m eigenvalue. To investigate this, we follow the same procedure, but with Jˆz: ˆ ˆ ˆ ˆ ˆ ˆ Jz(J j, m )=(J Jz ~J ) j, m = ~(m 1)(J j, m ). (1.38) ± | i ± ± ± | i ± ± | i We see that the m eigenvalue is increased (decreased) by 1 when acted on by Jˆ+ (Jˆ ). Putting together Eq. 1.37 and Eq. 1.38 we can state: Jˆ j, m = C (j, m) j, m 1 (1.39) ± | i ± | ± i where C (j, m) can be determined by taking the inner product of Eq. 1.39 with its Hermitian ± conjugate: 2 j, m 1 C⇤ (j, m)C (j, m) j, m 1 = C (j, m) , h ± | ± ± | ± i | ± | = j, m Jˆ Jˆ j, m , ⌥ ± (1.40) h | ˆ2 |ˆ2 i ˆ = j, m J Jz ~Jz j, m , h | ⌥ | i = ~2[j(j +1) m(m 1)]. ± This equation shows the convenience of using j(j +1)astheeigenvalueofJˆ2. While this equation is simple, we can use it to determine many things about our vector space. For one thing, since C (j, m) 2 is the norm of a vector, it must be non-negative, i.e. we must have ± j(j +1) m|(m 1) | 0. We can analyze the + and - cases separately. ± (+) j(j +1) m(m +1) 0= j 1 m j, (1.41) ( ) j(j +1) m(m 1) 0=)j m j +1. )  These conditions need to both be true at all times, and so we must use the strictest conditions: j m j. (1.42)   So far there are no requirements on j. However, let’s consider a state less than one unit below the maximum value of m, i.e. j, m k for some 0